Question about inserting a dielectric material inside a capacitor

In summary, the example discusses a parallel plate capacitor with a capacitance of 13.5 pF and a potential difference of 12.5 V between its plates. A porcelain blade is introduced between the plates, causing the final energy of the system to be less than the initial energy by a factor of 1/κe. This is due to the fact that the capacitor exerts a force on the blade, causing it to oscillate between the plates. The blade's oscillatory motion is parallel to the plates and its energy would be 893 pJ when it completely fills the space between the plates. However, in real life, part of the energy is consumed for polarizing the dielectric and is lost as heat and radiation
  • #1
pc2-brazil
205
3
I am self-studying "Physics" (Volume 3) by Halliday, Resnick and Krane, and would like to ask some questions regarding the following example given in the book (with solution). I will copy the entire example here, with solution (translated from Portuguese, because I am using a Portuguese translation of this book):

"A parallel plate capacitor, whose capacitance C0 is 13.5 pF, is subject to a potential difference of 12.5 V between its plates. The battery is turned off and a porcelain blade (κe = 6.5) is put between the plates (filling the space between the plates). What is the energy stored in the system, before and after the introduction of the porcelain?

The initial stored energy is given by [itex]U_i = \frac{1}{2}C_0V^2[/itex] = 1055 pJ.
The final energy is [itex]U_f = \frac{q^2}{2C} = \frac{q^2}{2\kappa_eC_0} = \frac{U_i}{\kappa_e}[/itex] = 162 pJ.

The final energy is less than the initial energy by a factor of 1/κe.
The "missing" energy is due to the fact that the capacitor exerted a force on the blade, doing work given by:
[itex]W = U_i - U_f[/itex] = 1055 pJ - 162 pJ = 893 pJ.
If the only force applied to the blade is the one exerted by the capacitor (ignoring all additional forces, such as friction), it will oscillate between the plates of the capacitor. The capacitor + blade system has a constant energy of 1055 pJ; the energy also oscillates between kinetic energy of the moving blade and energy stored in the electric field. In the instant in which the oscillating blade completely filled the space between the plates of the capacitor, its energy would be 893 pJ."

I would like to see if I understood the explanation above.
When it says "the capacitor exerted a force on the blade", does it mean that the blade is being attracted by the plates by electrostatic induction, that is, the capacitor is inducing a non-uniformly distributed charge on the blade during its insertion between the plates?
When it says "it will oscillate between the plates of the capacitor", does it mean that, if the plate is abandoned, partially inserted between the plates of the capacitor, it will perform an oscillatory movement parallel to the plates?

Thank you in advance.
 
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  • #2
The material of the blade will be electrostatically polarized.

Yes, the blade's oscillatory motion is parallel to the plates.
 
  • #3
In the real life, you will not see a blade oscillating forever between the plate of a capacitor. Part of the energy of the electric field is consumed for polarizing the dielectric. That means opposite charges moving in opposite directions in the molecules of the dielectric, removing them from their equilibrium positions and increasing their energy. So not all the work done by electric field of the capacitor transforms into kinetic energy of the dielectric, even in lack of friction. The work done on the molecules and dipoles is lost, as it transforms to heat and radiation.

ehild
 
  • #4
When it says that "The capacitor + blade system has a constant energy of 1055 pJ", does it mean that the total energy is initially the energy in the capacitor's electric field, and the blade has initially no energy because it is at rest?
So, the blade is pushed by the electric field and its maximum kinetic energy is 893 pJ and it occurs at the point when it fills entirely the space between the plates. On the other hand, if the blade is inserted into the capacitor by an external agent, this agent will have to do a negative total work (equal to -893 pJ) to insert the blade at constant velocity, by exerting in the blade a force of equal magnitude and opposite direction than the force that the capacitor is exerting to push the blade. Is this right?
 
  • #5


I can confirm that your understanding of the explanation is correct. When the capacitor is initially charged, it creates an electric field between the plates. When the dielectric material (in this case, the porcelain blade) is inserted between the plates, it is polarized by the electric field and becomes attracted to the plates. This attraction creates a force on the blade, causing it to move towards the plates.

As the blade is inserted between the plates, it experiences an oscillatory motion due to the changing electric field. This motion is parallel to the plates and is a result of the energy being transferred between the kinetic energy of the blade and the stored energy in the electric field. This oscillation will continue until the blade completely fills the space between the plates, at which point the energy in the system will be equal to the final energy calculated in the example.

I hope this helps to clarify your understanding of the example. Keep up the good work in your self-study of physics!
 

FAQ: Question about inserting a dielectric material inside a capacitor

What is a dielectric material?

A dielectric material is a non-conductive substance that is used to separate the plates inside a capacitor. It is usually made of an insulating material such as plastic, glass, or ceramic.

Why is a dielectric material necessary in a capacitor?

A dielectric material is necessary in a capacitor because it helps to increase the capacitance, or the ability to store electrical charge, of the capacitor. It also helps to prevent the flow of current between the plates, allowing the capacitor to hold its charge for a longer period of time.

How does a dielectric material affect the capacitance of a capacitor?

The dielectric material affects the capacitance of a capacitor by increasing it. This is because the material has a higher permittivity, or ability to store electrical energy, compared to air or a vacuum. This allows for more charge to be stored on the plates of the capacitor.

Can any material be used as a dielectric in a capacitor?

No, not all materials can be used as a dielectric in a capacitor. The material must have a high electrical resistivity and a low dielectric loss in order to effectively increase the capacitance of the capacitor. Some common materials used as dielectrics include paper, mica, and ceramic.

How does the thickness of the dielectric material affect the capacitance of a capacitor?

The thickness of the dielectric material affects the capacitance of a capacitor by directly influencing the distance between the plates. The closer the plates are, the higher the capacitance will be. Therefore, a thicker dielectric material will result in a higher capacitance compared to a thinner material.

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